A proton and an α -particle are accelerated from rest by 2 V and 4 V potentials, respectively. The ratio of their de-Broglie wavelength is: (a) 4: 1 (b) 2: 1 (c) 8: 1 (d) 16: 1

Answer: (a). λ = (h)/(√(2mqV)) ⇒ λ ∝ (1)/(√(mqV)). Proton: m · e · 2 = 2me. Alpha: 4m · 2e · 4 = 32me. (λp)/(λα) = √(32/2) = 4 So the answer is (a) 4:1.

A proton and an α -particle are accelerated from rest by 2 V and 4 V… | JEE Main 2023 (12 Apr, Shift 1)

Q75JEE Main 2023 · 12 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

A proton and an $\alpha$ -particle are accelerated from rest by $2\text{ V}$ and $4\text{ V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:

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Answer: (a)

$\lambda = \dfrac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mqV}}$ .
Proton: $m \cdot e \cdot 2 = 2me$ . Alpha: $4m \cdot 2e \cdot 4 = 32me$ .
$\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{32}{2}} = 4$
So the answer is (a) $4:1$ .

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